Volume of Sphere/Proof by Cavalieri

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Theorem

The volume $V$ of a sphere of radius $r$ is given by:

$V = \dfrac {4 \pi r^3} 3$


Proof

VolumeOfSphereCavalieri.png

Consider a sphere $S$ of radius $r$.

Consider a right circular cylinder $C$ whose bases are circles of radius $r$ and whose height is $2 r$.

Let a double napped cone $K$ each of whose nappes has bases which coincide with the bases of $C$.

Let $K$ be removed from $C$ to leave a solid figure $C'$ described as a right circular cylinder with a conical hollow at either end.


The bases of $C$ are circles of radius $r$.

From Area of Circle, the area of each base is therefore $\pi r^2$.

From Volume of Right Circular Cylinder, $C$ has volume given by:

\(\ds V_C\) \(=\) \(\ds \pi r^2 \times 2 r\)
\(\ds \) \(=\) \(\ds 2 \pi r^3\)


$K$ has twice the volume of each of its nappes.

Each of the nappes of $K$ is of height $r$ and has a base of area $\pi r^2$.

Hence from Volume of Cone, $K$ has volume given by:

\(\ds V_K\) \(=\) \(\ds 2 \times \dfrac 1 3 \times \pi r^2 \times r\)
\(\ds \) \(=\) \(\ds \dfrac 2 3 \pi r^3\)


Thus the volume $V_{C'}$ of $C'$ is given by:

\(\ds V_{C'}\) \(=\) \(\ds 2 \pi r^3 - \dfrac 2 3 \pi r^3\)
\(\ds \) \(=\) \(\ds \dfrac {\left({6 - 2}\right) \pi r^3} 3\)
\(\ds \) \(=\) \(\ds \dfrac {4 \pi r^3} 3\)


Let the sphere $S$ be positioned so as to be tangent to the plane containing the base of $C$

Consider a plane parallel to the base of $C$ intersecting both $S$ and $C$ at a distance $x$ from the center of $S$.

Let $y$ be the radius of the circle which is the intersection of $S$ with that plane.

From Pythagoras's Theorem, $y$ is given by:

$x^2 + y^2 = r^2$

and so:

$y^2 = r^2 - x^2$

Thus the area of this circle is given by:

$A = \pi \left({r^2 - x^2}\right)$


Similarly, consider the intersection of $C'$ with the same plane.

This consists of:

the circle of radius $r$ which has an area equal to that of the base of $C$

with:

the circle of radius $x$ which is the intersection of the plane with the cone

removed from it.

So the area of the remaining circle with the inner circle removed is:

$A' = \pi r^2 - \pi x^2$

from Area of Circle.


Thus the intersection of the plane with both $S$ and $C'$ are the same.

Thus by Cavalieri's Principle $S$ and $C'$ have the same volume.

Hence the result.

$\blacksquare$


Historical Note

This proof was given by Bonaventura Francesco Cavalieri.


Sources

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